In trapezoid $ABCD$, $AB=BC=2AD$ and $AD=5$ units. What is the area of trapezoid $ABCD$? [asy] draw((0,0)--(10,0)--(10,10)--(0,5)--cycle,black+linewidth(1)); draw(rightanglemark((0,5),(0,0),(10,0),20),black+linewidth(1)); draw(rightanglemark((0,0),(10,0),(10,10),20),black+linewidth(1)); label("A",(0,0),W); label("B",(10,0),E); label("C",(10,10),E); label("D",(0,5),W); [/asy]
Solution: Since $AB$ and $BC$ are twice the length of $AD$ and $AD=5$ units, $AB=BC=2\cdot5=10$ units. The area of a trapezoid is given by $\frac{b_1+b_2}{2}h$, where $b_1$ and $b_2$ are the bases of the trapezoid and $h$ is the height of the trapezoid. In trapezoid $ABCD$, $AB$ is the height and $AD$ and $BC$ are the bases. Therefore, the area of trapezoid $ABCD$ is \begin{align*} \frac{BC+AD}{2}(AB)&=\frac{10\text{ units}+5\text{ units}}{2}\cdot10\text{ units}\\&=\left(\frac{15}{2}\text{ units}\right)\left(10\text{ units}\right)\\&=\boxed{75}\text{ units}^2.\\\ \end{align*}